Question Let $g\in C_0^\infty((-1,1))$.Prove $\forall t\in (-1,1)$,$${g^4}\left( t \right) \le 16\int_{ - 1}^1 {\left( {{{\left| {g'\left( s \right)} \right|}^2} - \frac{{{g^2}\left( s \right)}}{{4{{\left( {1 - \left| s \right|} \right)}^2}}}} \right)ds} \cdot \int_{ - 1}^1 {{{\left| {g\left( s \right)} \right|}^2}ds} .$$
Theorem. Let $g \in C_0^\infty((-1,1))$. Then for all $t \in (-1,1)$, $$g^4(t) \le 16 \int_{-1}^1 \left( |g'(s)|^2 - \frac{g^2(s)}{4(1 - |s|)^2} \right) ds \int_{-1}^1 |g(s)|^2 ds$$
Proof. Let $h(s) = g^2(s)$. \begin{align*} \left( \int_{-1}^1 \left( |g'(s)|^2 - \frac{g^2(s)}{4(1 - |s|)^2} \right) ds \right)^2 &\le \int_{-1}^1 ds \int_{-1}^1 |g'(s)|^4 - \frac{g^2(s) |g'(s)|^2}{2(1 - |s|)^2} ds \\ &= \int_{-1}^1 |g'(s)|^4 ds - \frac{1}{2} \int_{-1}^1 \frac{g^2(s) |g'(s)|^2}{(1 - |s|)^2} ds \\ &\le \int_{-1}^1 |g'(s)|^4 ds - \frac{1}{2} \int_{-1}^1 \frac{g^4(s)}{(1 - |s|)^4} ds. \end{align*}By the AM-GM inequality, \begin{align*} \frac{g^2(s) |g'(s)|^2}{(1 - |s|)^2} &\ge 4 \left( \frac{|g'(s)|}{2(1 - |s|)} \right)^4 \\ &= 4 \left( \frac{g^2(s)}{4(1 - |s|)^2} \right)^2 \\ &= \frac{g^4(s)}{(1 - |s|)^4}. \end{align*} $\int_{-1}^1 |g'(s)|^4 ds \le 2 \int_{-1}^1 g^4(s) ds$ \begin{align*} $$\int_{-1}^1 g^4(s) ds \int_{-1}^1 |g(s)|^2 ds &\ge \left( \int_{-1}^1 g^2(s) ds \right)^2 &= h^2(-1) \\ &= g^4(-1). \end{align*}Finally, by the AM-GM inequality, $$\sqrt{2 g^4(-1) \int_{-1}^1 |g(s)|^2 ds} \le \sqrt{2} \sqrt{g^4(-1)} \sqrt{\int_{-1}^1 |g(s)|^2 ds} = 2g^2(-1).$$